C#中对象的不同列表
我必须区分对象列表,但不仅仅是ID,因为有时两个不同的对象具有相同的ID。 我有课:
public class MessageDTO { public MessageDTO(MessageDTO a) { this.MsgID = a.MsgID; this.Subject = a.Subject; this.MessageText = a.MessageText; this.ViewedDate = a.ViewedDate; this.CreatedDate = a.CreatedDate; } public int? MsgID { get; set; } public string Subject { get; set; } public string MessageText { get; set; } public System.DateTime? ViewedDate { get; set; } public System.DateTime? CreatedDate { get; set; } }
我怎么能区分清单:
List example;
谢谢
使用LINQ。
public class MessageDTOEqualityComparer : EqualityComparer { public bool Equals(MessageDTO a, MessageDTO b) { // your logic, which checks each messages properties for whatever // grounds you need to deem them "equal." In your case, it sounds like // this will just be a matter of iterating through each property with an // if-not-equal-return-false block, then returning true at the end } public int GetHashCode(MessageDTO message) { // your logic, I'd probably just return the message ID if you can, // assuming that doesn't overlap too much and that it does // have to be equal on the two } }
然后
return nonDistinct.Distinct(new MessageDTOEqualityComparer());
您还可以通过重写object.GetHashCode()
object.Equals(object)
和object.GetHashCode()
并调用object.GetHashCode()
的空重载来避免需要额外的类。 但是,请确保您认识到此决策的含义:例如,这些将成为所有非显式范围内的相等测试函数。 这可能是完美的,正是您所需要的,或者可能导致一些意想不到的后果。 只要确保你知道你正在进入什么。
我想使用其他属性,你应该实现IEqualityComparer
接口。 更多信息: msdn
class MsgComparer : IEqualityComparer { public bool Equals(MessageDTO x, MessageDTO Oy) { } // If Equals() returns true for a pair of objects // then GetHashCode() must return the same value for these objects. public int GetHashCode(MessageDTO m) { //it must br overwritten also } }
然后:
example.Distinct(new MsgComparer());
您还可以在MessageDTO
类中覆盖 Equals
:
class MessageDTO { // rest of members public override bool Equals(object obj) { // your stuff. See: http://msdn.microsoft.com/en-us/library/ms173147%28v=vs.80%29.aspx } public override int GetHashCode() { } }
那就够了:
example.Distinct();
您可以使用MoreLinq库中的扩展方法DistinctBy:
string[] source = { "first", "second", "third", "fourth", "fifth" }; var distinct = source.DistinctBy(word => word.Length);
看到这里 :
我建议你使用@Matthew Haugen的解决方案
如果您不想为此创建新类,可以通过将不同字段的列表分组来使用LINQ,然后选择该组中的第一个项目。 例如:
example.(e => new { e.MsgID, e.Subject }).Select(grp => grp.FirstOrDefault());